Reduce series, parallel, and feedback blocks to one transfer function, then read off closed-loop gain, sensitivity, and disturbance rejection.
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B. Bode plots
Bode Plots
Build straight-line magnitude and phase asymptotes from poles and zeros, then read gain margin, phase margin, and stability straight off the plot.
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C. Closed-loop response, open-loop response, and stability
Closed-Loop Response and Stability
Find pole locations from the characteristic equation, test stability with Routh-Hurwitz and root locus, and tie BIBO stability to margins.
Stability is the first question any control system must answer: will a bounded input produce a bounded output, or will the response blow up? Everything reduces to where the closed-loop poles sit in the complex s-plane. The FE handbook states the rule that anchors this whole topic — the closed-loop characteristic equation is 1+G1(s)G2(s)H(s)=0, and its roots are the poles whose locations decide stability. This concept is the storefront for the controls unit because nearly every C-area question is a disguised pole-location problem.
The characteristic equation
For the loop 1+G(s)H(s)=0, clearing fractions gives a polynomial in s whose roots are the closed-loop poles. Open-loop poles are the roots of the forward/loop transfer function's denominator; closing the loop moves them. Set the denominator of T(s)=G/(1+GH)
Pole locations and BIBO stability
A continuous LTI system is bounded-input bounded-output (BIBO) stable if and only if every closed-loop pole has a strictly negative real part — all poles in the open left half-plane. A pole on the imaginary axis (Re=0) gives a marginally stable, sustained oscillation; any pole in the right half-plane gives an unbounded, growing response. Complex poles −σ±jωd produce a decaying (σ>0
Routh-Hurwitz criterion
Routh-Hurwitz tells you how many roots lie in the right half-plane without factoring the polynomial. Build the Routh array from the characteristic polynomial ansn+⋯+a0: the first two rows hold alternating coefficients, and each lower entry is the negative of the 2×2
Finding the marginal gain
The most common Routh question gives a gain K inside the characteristic polynomial and asks for the range of K that keeps the system stable, or the value of K at the edge of instability. Force a first-column entry to zero to find the marginal gain; the auxiliary equation formed from the row above that zero then gives the frequency of the sustained oscillation. This is the algebraic twin of a pole crossing the imaginary axis.
first-column entry=0⇒Kmarg,A(s)=0⇒ωosc
Root locus
The root locus traces how the closed-loop poles migrate in the s-plane as a gain K sweeps from 0 to ∞. The locus starts at the open-loop poles (K=0) and ends at the open-loop zeros or at infinity (K→∞
Gain margin and relative stability
Being merely stable is not enough; you want margin. The linear gain margin is the factor between the current gain and the marginal gain found above, GMlin=Kmarg/Kdesign — a bare, unitless ratio. The handbook's dB gain margin is
Exam strategy
First form the characteristic equation by setting 1+GH=0 and clearing fractions — never test the open-loop denominator. For a numeric polynomial, scan for missing or negative coefficients (instant instability) before building the Routh array. When K appears, write the array symbolically, set the first-column term to zero for Kmarg, and use the auxiliary equation for the oscillation frequency. Reserve root-locus reasoning for 'which way do the poles move' questions. Keep BIBO crisp: strictly-left-half-plane poles only — a pole on the
Problem. A unity-feedback system has open-loop transfer function G(s)=s3+2s2+sK
Common pitfalls
•Testing the open-loop denominator for stability. You must form the closed-loop characteristic equation 1+GH=0 and clear fractions first; the open-loop poles are not the closed-loop poles.
•Treating a pole on the imaginary axis as stable. Re(p)=0 is marginally stable (sustained oscillation), not BIBO stable, which requires strictly Re(p)<0
References
NCEES FE Reference Handbook — Instrumentation, Measurement, and Control: Control Systems — Closed-loop characteristic equation $1+G_1G_2H=0$, Final Value Theorem dc gain, and gain/phase-margin definitions.
NCEES FE Reference Handbook — Electrical and Computer Engineering: Decibels and Bode Plots — Frequency-response basis for the gain-margin computation.
Ogata, Modern Control Engineering — Routh-Hurwitz array construction and root-locus rules (off-handbook supplements for area C), ch. 5-6.
D. Controller performance
Controller Performance and PID
Compute steady-state error from system type and error constants, read transient specs off the damping ratio, and tune P/PI/PID actions.
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to zero — that is the characteristic equation, and its roots, not the open-loop poles, govern the closed-loop response.
1+G(s)H(s)=0⟺denominator of T(s)=0
) or growing (
σ<0
) sinusoid; the real part sets the envelope and the imaginary part sets the ringing frequency.
BIBO stable⟺Re(pi)<0for all poles pi
determinant of the two rows above, divided by the pivot (equivalently, cross-multiply in the order shown in the equation below). The system is stable if and only if every entry in the first column is present and has the same sign; the number of sign changes equals the number of right-half-plane roots. A necessary first check: all coefficients must be present and positive.
bi=an−1an−1an−2i−anan−2i−1
). A branch lies on the real axis to the left of an odd count of real poles and zeros. The system goes unstable at the gain where a branch crosses the imaginary axis — exactly the marginal
K
Routh finds. Root locus is the picture; Routh is the calculation; they agree at the crossing.
, read off the Bode magnitude at the phase-crossover frequency
ω180
. Keep the two forms distinct: a linear factor of
5
corresponds to
20log105≈14dB
, not
5dB
, so never set the bare ratio equal to the dB value. A large gain margin means the poles sit well inside the left half-plane; a small one means a modest gain increase (or modeling error) could push them across. Relative stability is quantified by both gain and phase margins.
Each entry is the negative of the 2×2 determinant of the two rows above, divided by the pivot an−1 — i.e. cross-multiply in the order shown in the latex (pivot-row column term minus top-row column term). Continue until the s0 row.
Routh stability teststable⟺first column all same sign
Number of first-column sign changes = number of right-half-plane roots.
Formed from the row above a zeroed first-column entry; gives the sustained-oscillation frequency at Kmarg.
Root-locus condition1+KD(s)N(s)=0
Loci start at open-loop poles (K=0), end at open-loop zeros or infinity (K→∞).
Gain margin (dB)GMdB=−20log10G(jω180)=20log10KdesignKmarg
Handbook dB form, evaluated at the phase-crossover ω180. The linear gain margin is the bare factor Kmarg/Kdesign; the dB value is 20log10
, giving the characteristic equation
1+G(s)=0⇒s3+2s2+s+K=0
. Find the range of
K
for stability and the oscillation frequency at the stability limit.
Solution. Routh array for s3+2s2+s+K:
s3:1,1s2:2,Ks1:2(2)(1)−(1)(K)=22−K,0s0:K
First column all positive requires 22−K>0 and K>0, so 0<K<2.
Marginal at K=2: auxiliary equation from the s2 row is 2s2+K=2s2+2=0⇒s2=−1⇒ωosc=1rad/s.
Sanity check: at K=2 a pole pair sits on the jω axis at ±j1, the boundary between stable and unstable — consistent with sustained 1rad/s oscillation.
s1row:22−K>0⇒0<K<2
Count right-half-plane roots
Problem. Is the system with characteristic polynomial s4+2s3+3s2+4s+5=0 stable? If not, how many unstable poles?
Closed-loop poles from the characteristic equation
Problem. A unity-feedback loop has G(s)=s+28. Find the closed-loop pole and state whether the system is BIBO stable.
Solution. Characteristic equation: 1+G(s)=0⇒1+s+28=0⇒(s+2)+8=0⇒s=−10.
The single closed-loop pole is at s=−10, strictly in the left half-plane.
BIBO stable. Sanity check: the open-loop pole at s=−2 moved left to s=−10 when the loop closed (faster response), and the dc gain became T(0)=8/10=0.8 — both consistent with high-gain negative feedback.
1+s+28=0⇒s=−10(stable)
.
•Concluding stability from all-positive coefficients. That is necessary but not sufficient for order ≥3; the fourth-order example passes the coefficient test yet has two RHP roots.
•Sign-error in the Routh determinant. The numerator is (pivot row times the column term above-right) minus (top entry times the next term) — i.e. the negative of the literal 2x2 determinant; divide by the pivot, and keep the cross-multiplication order consistent.
•Conflating the linear and dB gain margins. The bare factor Kmarg/Kdesign is the linear gain margin; the dB margin is 20log10 of that ratio (a factor of 5 is about 14 dB, not 5 dB). Never equate the unitless ratio to the −20log10∣G(jω180)∣ value.
•Reading the root locus backward: branches begin at open-loop poles and terminate at open-loop zeros (or infinity) as K increases, not the reverse.
of that ratio (e.g. a factor of
5≈14
dB). Positive dB = stable, minimum-phase.
s1:1(1)(4)−(2)(5)=−6
s0:5
First column:
1,2,1,−6,5
. Sign goes
+→+→+→−→+
: two sign changes.
Two right-half-plane roots, so the system is unstable.
Sanity check: all coefficients are positive, so the necessary condition passes — but Routh exposes the two RHP roots that the coefficient test alone would miss.